# Thread: i forget a very simple integration problem

1. ## i forget a very simple integration problem

i forget a very simple integration problem: is there any way to do this:

2. $\int {\sec \theta \,d\theta } = \int {\frac{{\cos \theta }}
{{(1 + \sin \theta )(1 - \sin \theta )}}\,d\theta } = \frac{1}
{2}\left( {\int {\frac{{\cos \theta }}
{{1 + \sin \theta }}\,d\theta } + \int {\frac{{\cos \theta }}
{{1 - \sin \theta }}\,d\theta } } \right).$

Or just multiply top & bottom by $\sec\theta+\tan\theta.$

3. Hello, afeasfaerw!

$\int \sec\theta\,d\theta$

You should have been taught the formula for this by now . . .

It is usually derived like this . . .

Multiply by . $\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}$

. . $\int\frac{\sec\theta}{1}\cdot\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\,d\theta \;=\;\int\frac{\sec^2\!\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\,d\theta \;=\;\int\frac{\sec\theta\tan\theta + \sec^2\!\theta}{\sec\theta + \tan\theta}\,d\theta$

Let: $u \:=\:\sec\theta + \tan\theta\quad\Rightarrow\quad du \:=\:(\sec\theta\tan\theta + \sec^2\theta)d\theta$

Substitute: . $\int\frac{du}{u}\;=\;\ln|u| + C$

Back-substitute: . $\ln|\sec\theta + \tan\theta| + C$